Explanation Given information: F ( x ) is an antiderivative of f ( x ) = 3 e − x 2 and F ( 0 ) = 5 Formula used: Fundamental theorem of calculus :If f is continuous on the closed interval [ a , b ] and F is the indefinite integral of f on [ a , b ] , then ∫ a b f ( x ) d x = F ( b ) − F ( a ) Calculation: F ( x ) is an antiderivative of f ( x ) = 3 e − x 2 and F ( 0 ) = 5 (Given) ∫ 0 2 f ( x ) d x = ∫ 0 2 3 e − x 2 d x = 3 ∫ 0 2 e − x 2 d x = 3 ∫ 0 2 e − t d t 2 t [ l e t x 2

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 6.1, Problem 4P

To determine

F(2) if F(x) is an antiderivative of f(x)=3e−x2 and F(0)=5.

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Chapter 6.1, Problem 3P

Chapter 6.1, Problem 5P

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on donne f(x) da fonction derive dhe do fonction fcsos calcule f'(x) orans chacun des Cas sulants: 3 1) f(x)=5x-11, 2- f (x) = ->³ 3-1(x) = x² 12x +π; 4-f(x)=- 5-f(x) = 33-4x6-609)=-3x²+ 7= f(x) = x + 1.8-f(x) = 4 s-f(x) = x++ X+1 -x-1 2 I 3x-4 дево

The correct answer is Ccould you show me how to do it by finding a0 and and akas well as setting up the piecewise function and integrating

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Applied Calculus, 6e WileyPLUS + Loose-leaf

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F ( 2 ) if F ( x ) is an antiderivative of f ( x ) = 3 e − x 2 and F ( 0 ) = 5 .

Solution Summary: The author calculates that F(2) is an antiderivative of f(x)=3e-x2 and F(0)=5.

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F(2) if F(x) is an antiderivative of f(x)=3e−x2 and F(0)=5.

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Chapter 6 Solutions